Question

我注意到，scipy.special秩序n和参数x jv(n,x)的贝塞尔函数在x中被矢量化：

In [14]: import scipy.special as sp In [16]: sp.jv(1, range(3)) # n=1, [x=0,1,2] Out[16]: array([ 0., 0.44005059, 0.57672481])

但是没有相应的矢量化形式的球形贝塞尔函数，sp.sph_jn：

In [19]: sp.sph_jn(1,range(3)) 

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-19-ea59d2f45497> in <module>()
----> 1 sp.sph_jn(1,range(3)) #n=1, 3 value array

/home/glue/anaconda/envs/fibersim/lib/python2.7/site-packages/scipy/special/basic.pyc in sph_jn(n, z)
    262     """
    263     if not (isscalar(n) and isscalar(z)):
--> 264         raise ValueError("arguments must be scalars.")
    265     if (n != floor(n)) or (n < 0):
    266         raise ValueError("n must be a non-negative integer.")

ValueError: arguments must be scalars.

此外，球形贝塞尔函数一次计算N的所有阶数。因此，如果我想要参数n=5的{{1}}贝塞尔函数，则返回n = 1,2,3,4,5。它实际上在一次传递中返回jn及其衍生物：

x=10

为什么这种不对称存在于API中，并且有没有人知道一个库将返回矢量化的球形贝塞尔函数，或者至少更快（即在cython中）？

Answer 1

您可以编写一个cython函数来加速计算，首先要做的是获取fortran函数SPHJ的地址，这里是如何在Python中执行此操作：

from scipy import special as sp
sphj = sp.specfun.sphj

import ctypes
addr = ctypes.pythonapi.PyCObject_AsVoidPtr(ctypes.py_object(sphj._cpointer))

然后你可以直接在Cython中调用fortran函数，注意我使用prange()来使用多核来加速计算：

%%cython -c-Ofast -c-fopenmp --link-args=-fopenmp
from cpython.mem cimport PyMem_Malloc, PyMem_Free
from cython.parallel import prange
import numpy as np
import cython
from cpython cimport PyCObject_AsVoidPtr
from scipy import special

ctypedef void (*sphj_ptr) (const int *n, const double *x, 
                            const int *nm, const double *sj, const double *dj) nogil

cdef sphj_ptr _sphj=<sphj_ptr>PyCObject_AsVoidPtr(special.specfun.sphj._cpointer)


@cython.wraparound(False)
@cython.boundscheck(False)
def cython_sphj2(int n, double[::1] x):
    cdef int count = x.shape[0]
    cdef double * sj = <double *>PyMem_Malloc(count * sizeof(double) * (n + 1))
    cdef double * dj = <double *>PyMem_Malloc(count * sizeof(double) * (n + 1))
    cdef int * mn    = <int *>PyMem_Malloc(count * sizeof(int))
    cdef double[::1] res = np.empty(count)
    cdef int i
    if count < 100:
        for i in range(x.shape[0]):
            _sphj(&n, &x[i], mn + i, sj + i*(n+1), dj + i*(n+1))
            res[i] = sj[i*(n+1) + n]    #choose the element you want here        
    else:
        for i in prange(count,  nogil=True):
            _sphj(&n, &x[i], mn + i, sj + i*(n+1), dj + i*(n+1))
            res[i] = sj[i*(n+1) + n]    #choose the element you want here
    PyMem_Free(sj)
    PyMem_Free(dj)
    PyMem_Free(mn)
    return res.base

比较一下，这是在forloop中调用sphj()的Python函数：

import numpy as np
def python_sphj(n, x):
    sphj = special.specfun.sphj
    res = np.array([sphj(n, v)[1][n] for v in x])
    return res

以下是10个元素的％timit结果：

x = np.linspace(1, 2, 10)
r1 = cython_sphj2(4, x)
r2 = python_sphj(4, x)
assert np.allclose(r1, r2)
%timeit cython_sphj2(4, x)
%timeit python_sphj(4, x)

结果：

10000 loops, best of 3: 21.5 µs per loop
10000 loops, best of 3: 28.1 µs per loop

以下是100000个元素的结果：

x = np.linspace(1, 2, 100000)
r1 = cython_sphj2(4, x)
r2 = python_sphj(4, x)
assert np.allclose(r1, r2)
%timeit cython_sphj2(4, x)
%timeit python_sphj(4, x)

结果：

10 loops, best of 3: 44.7 ms per loop
1 loops, best of 3: 231 ms per loop

Answer 2

如果有人仍然感兴趣，我发现一个解决方案比Ted Pudlik的解决方案快了近17倍。我使用了这样一个事实，即n阶球面贝塞尔函数基本上是n + 1/2阶标准贝塞尔函数的1 / sqrt（x）倍，它已经被矢量化了：

import numpy as np
from scipy import special

sphj_bessel = lambda n, z: special.jv(n+1/2,z)*np.sqrt(np.pi/2)/(np.sqrt(z))

我得到了以下时间：

%timeit sphj_vectorize(2, x) # x = np.linspace(1, 2, 10**5)
1 loops, best of 3: 759 ms per loop

%timeit sphj_bessel(2,x) # x = np.linspace(1, 2, 10**5)
10 loops, best of 3: 44.6 ms per loop

Answer 3

有一个pull request将矢量化球形Bessel函数例程合并到SciPy中scipy.special.spherical_x，x = jn, yn, in, kn。运气不错，他们应该把它变成版本0.18.0。

np.vectorize以上的性能提升（即for-loop）取决于函数，但可能是数量级。

import numpy as np
from scipy import special

@np.vectorize
def sphj_vectorize(n, z):
    return special.sph_jn(n, z)[0][-1]

x = np.linspace(1, 2, 10**5)

%timeit sphj_vectorize(4, x)
1 loops, best of 3: 1.47 s per loop

%timeit special.spherical_jn(4, x)
100 loops, best of 3: 8.07 ms per loop

python中的矢量化球形bessel函数？

3 个答案: